Question

If $\alpha$ is a real root of $2x^3-3x^2+6x+6=0$, then find $\left[\alpha \right]$ where $[\cdot ]$ denotes the greatest integer function.

Answer 1

If $\alpha$ is the root of ${2x}^3-{3x}^2+6x+6=0$ find $\left[\alpha \right].$

$\because {2x}^3-{3x}^2+6x+6=0$

$f^{\prime }\left(x\right)$ $=\frac{d}{dx}({2x}^3-{3x}^2+6x+6)=x^2-x+6=6(x^2+x-1)$

$\because$ Discriminent of $f^{\prime }\left(x\right)<0$

Hence $f^{\prime }\left(x\right)$ is always greater than $0.$

Hence $f\left(x\right)$ is an increasing function, which means that it has only one root.

Let $x=-1$

$f(-1)=2{(-1)}^3-3{(-1)}^2+6(-1)+6=-2-3-6+6=-5$

Let $x=0$

$f(-1)=2{\left(0\right)}^3-3{\left(0\right)}^2+6\left(0\right)+6=6$

Hence $f(-1)f\left(0\right)<0$

Which means that $\alpha$ should be in between $-1$ & $0.$

Hence $\left[\alpha \right]=-1$